PLease Help With Physics Question.?

Question by PhysicsKid12: PLease Help with physics Question.?
One hundred grams of ice at -10*C are placed in a 60 g aluminum calorimeter cup (assume the combination is in thermal equilibrium). Twenty grams of steam are injected into the ice, what is the final temperature of the combination?

Best answer:

Answer by Pearlsawme
Latent heat of ice = 80 cals/g
Latent heat of steam = 540cals/g

Specific heat capacity of ice = 0.5 cals /(g˚C)
Specific heat capacity of water = 1 cals /(g˚C)
Specific heat capacity of allulminium 0.215 cals /(g˚C)
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Assuming the temperature of steam to be at 100˚C
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Heat gained by ice of 100 g till it is changed into water at 0˚C = 100[80 + 0.5*15}
= 8750 cal

Heat gained by water of 100 g at 0*C till it reaches t˚C = 100 *t cals

Heat gained by aluminium of 60 g = 60 *0.215* (t + 15)
12.9 t + 193.5

Total heat gained = 8750 + 100 *t + 12.9t +193.5 cals
=8943.5+ 112.9 t cals.

Heat lost by steam 20*540 =10800 cals
Heat lost by water converted from steam 20* (100 -t) = 2000 -20t
Total heat lost 10800 + 2000 -20t

Heat lost = heat gained
10800 + 2000 -20t =8943.5+ 112.9 t cals.
3856.5 = 132.9

t =29. ˚C
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